takeshi yajima- on simply knotted spheres in r^4

8/3/2019 Takeshi Yajima- On Simply Knotted Spheres in R^4

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Yajima, T.Osaka J. Math.1 (1964), 133-152

ON SIMPLY KNOTTED SPHERES IN R4

Dedicated to Professor H. Terasaka on his 60-th birthday

T A K E S H I Y A J I M A

(Received September 30, 1964)

1. Introduction

Let S be a 2-sphere in the 4-dimensional Euclidean space R\ and

let S* be the projection of S on a 3-dimensional subspace. It is obviousthat the singularity of S* consists of several closed curves under appro-

priate conditions. A simplest case of the projection may be such a casethat the singularity of S* consists of several disjoint simple closed curves.W e call such a projection a simple projection. No w, is it possible to

give a projection of every 2-sphere in I?4 as a simple projection on some

3-subspace?It is the purpose of this paper to prove that there exist some kinds

of locally flat 2-spheres, each sphere of which can not be deformed sothat it has a simple projection. For this purpose we consider two kindsof spheres, simply knotted spheres and symmetric ribbon sheres, andprove that these kinds of spheres coincide. S pheres of the former aredefined as equivalent classes of spheres, each of which contains a sphere

S such that its projection S* on some 3-subspace is a simple projection,and spheres of the latter are known

15as a simple case of knotted

2-spheres in R4.

The sphere used in the proof is the one discovered by R. H . Fox 2),and a slight modification of H. Terasaka's result [8] enable us to provethat the Fox's sphere is distinct from a symmetric ribbon sphere.

2. Preliminaries

In this section we shall state some fundamental concepts and termi-nologies, most of which were explained in [10], concerning the projectionmethod for the knot theory in the 4-space. Throughout this paperterminologies are used in the semi-linear point of view.

1) [7], P. 135, E xam ple 10, 11 and [10], p. 70, Exa mp l e 1, 2.

2) [7], p. 136, E xam ple 12.



134 T. Y A J I M A

Let /?4 be the 4-dimensional Euclidean space with a coordinate system

(x , y, z, u). The 3-dimensional subspace of R * defined by u=Q is denoted

by R3. The half spaces of R

4 defined by u^O or u^O are denoted

by H + or H- respectively. With every point P=(x,y,z,u) of R\ we

associate the point P* =(x , y, z, 0) and the coordinate u= u(P\ We call P*

the trace or the projection of P, and u the height of P respectively. The

mapping π :P-^P* is called the projection as usual.

Let M be a 2-dimensional manifold in P4

with or without boundaries.

There is no loss of generality to assume the fo l lowing condi t ion:

(2.1) // P!, " ,PW are vertices of M, then the system of points(Pf, • • • ,P % ) is in general position in R

3.

It is natural to denote the set of traces of points of M by M*. In

virtue of (2.1), we can suppose that the set of cutting points of M*,

that is, the set of points of intersections of different 2-simplexes of M*,

constitutes a 1-dimensional complex. We denote the set of cutting points

of M* by Γ(M*).

We can also assume the fo l lowing conditions:

(2.2) A segment of Γ(M*) is the intersection of just two 2-simplexes.

(2. 3) There exist just three 2-simplexes through a double point ofΓ(M*).

From the same argument as in Fig. 1 of [10], we can suppose that:

(2. 4) / X M * ) consists of the following three kinds of polygons,

(1) closed polygons,

(2) polygonal arcs, whose endpoints belong to the projection of bounda-

ries of M ,(3) polygonal arcs, whose endpoints contain some singular cutting

points.

The definition of singular cutting points was stated in [10], but we donot need th e definition in this paper.

Let 7* be a cutting of / X M * ), namely a polygon of a kind of (1) or (2),

an d 7j, γ2

be inverse images of 7* on S by the projection. If u(P^>u(P^

fo r some points P γ P2€γ2

such that P % = P $ , then it is obvious that

w(Qi)>w(Q 2) for every pair of points Q i G γ ^ Q 2 G γ 2 such that Q f = Q f.

W e denote this situation simply by u(71)^>u(

ry

2).

The terminology of cutting is used sometimes for a continuous image

of a 2-manif old under the same conditions as (2.1), (2. 2) and (2. 3).

The presentation of the knot group 8(M) in the projection methodis as follows. Let J??, • • ,lτf be components of M*— Γ(M*), where Mis an arbitrary 2-dimensional closed manifold in R

4. Let 7* be the



S IMP LY KNO TTE D S P HE R E S 135

intersection of two surfaces Af =Σt[)2t*

land Af=Σf(JΣ

J*

l(Fig.1),

and 7* >7, be the inverse images of 7* on At

and As

respectively. Ifw(7ί)<X7/) then we call A

{the under surface and A, the over surface

respectively.

Fig. 1.

We use the notation in Fig. 1 to represent the relation between the

heights of two surfaces, where the small vector on the under surface

indicates the orientation of the over surface.

According to the results of [10] we have the presentation of S(M),

with generators σ, corresponding to each Σ f ( / = !,-••, k\ as follows:

Generators : ( < r1 9

•••, σk)

D efining relations: σy+1 =σ

y, σ

i+1=σjσiσj

l

for each 7*, where Σj\jΣJ+l

is the over surface and the

orientation vector points f rom Σ f to Σ&.

3. Symmetric ribbon spheres

First of all we shall state the definition of ribbons. Let D 0 be the

unit disk #2+/<^l, and let / be a continuous mapping of Z)0

into R * .

Put D*=/(I>0). Let 7* be a simple arc of Γ(D*), which is of the kind of

(2.4), (2), and is separated f rom the rest of Γ(Z)*). Then there happen

fo l lowing two cases (Fig.2):

(a) f~\y*) consists of two arcs on D0

such that the endpoints of

one of them belong to the boundary of D 0, an d those of the other do not.(b) /-1(7*) consists of two arcs on D

0such that one of the endpoints

of each arcs belongs to the boundary of D 0, and the others donot.



136 T. Y A J I M A

(a)

(b)

Fig. 2.

A ribbon τ is defined as a continuous image of D 0 in R3

such thatconsists of several disjoint simple arcs which are of the kind (a).

(3.1) Let τ be a ribbon. There exist disks D+(τ)czH*

+and D.(τ)

i such that their boundaries are contained in R* and D ! j? ( ϊ)=.D *(x)=x.

Proof. First of all we shall construct D+(τ). Let γf, • • • ,7* be cut-

tings of Γ(r) and let 7Ϊ, j* (y = l, ~ ,n) be the inverse image of 7*,

where th e endpoints of γj's are on the boundary of Z >0. Suppose F ϊ G γ J ,

PI 97? be the inverse image of the variable point P* on 7*. Let u be

a continuous function defined on Σ 7Ϊ U Σ 7? U -B^ A such( i ) if P* is an endpoint of 7*, then w ( P ϊ ) = 0 , w (

(ii) if P*G7* is not an endpoint of 7?, then 0<w(Pϊ)<w(P"),

(iii) if P* 6 5rf x-Σ 7?, then «(P)=0.

We can extend the function u onto D 0 such that u is continuous on D0.

It is obvious that the correspondence

h: P->(P*, w(P)) , P£ A

forms a homeomorphic mapping of D 0 into R *. Hence D+(τ)=h(D

0) is a

required disk.For the construction of £>_(r) it is sufficient to replace th e conditions

(i), (ii) by the following conditions:



S I M P L Y K N O T T E D S P H E R E S 137

(i') if P* is an endpoint of 7*, then u ( P l ) = Q ,

(iiO if P*G7* is not an endpoint of 7*, thenIt is obvious that:

(3.2) There exists a unique equivalent class of spheres which contains

the sphere S(τ) =D+(τ)(}D-(τ) for a given ribbon x.

We call this equivalent class of spheres the symmetric ribbon sphere

of r.It is easy to get the projection S* f rom an arbitrarily given ribbon

r. Let 7? be a cutting of x, and let ^4*, B* be parts of r which contain

7?, where B* runs through A*. Thicken r to a singular cube K*, so

that the ribbon knot k of r separates the surface S* of K* into twosingular disks, one of which corresponds to JOJ(x) and the other to D ΐ (x ) .Let At B* be the corresponding parts of 4*,5* in D * ( x ) and let A*, B*

be the parts in D *(x). Then the cutting 7? becomes to closed cuttings

7ft =>l*n(βϊU£ϊ) and γΐ2

=A!ί{}(B$θB*). By these cuttings the tubeB^iJBΐ. splits into three parts of tubes Tf, Γf, T

2*, and new disks C J

an d C* appear on A$ and ^4ΐ respectively (Fig. 3).

Fig. 3.

It is easily seen that 77 U Tf is the over surface at 7ft and that

T* U^*isthe u nd er surface at j f

2. Hereafter we say that A andB

have the opposite relations in heights at 7ft and 7?2. If we suppose that

the orientation of D?(t) coincides with that of & , then we have essential

generators σA, σ

fl l, σβ2 and the relation σ

Bl=σ

Aσ

B2σ-A\ where σ

A, σ

Bland

σB2

correspond to the surfaces A$(JA*, Tf()T$ and Tf respectively.

E X A M P L E 1. Let X i be a ribbon in Fig. 4, (a). The ribbon knot k ±

of X j is known as the square knot. The corresponding symmetric ribbon

sphere Sf is shown in (b).



138 T. Y A J I M A

Fig. 4.

EXAMPLE 2. Let x2

be a ribbon in Fig. 5, (a). The ribbon knot k2

of

r2

is known as the stevedore's knot.

( a ) (b )

Fig. 5.

Since the stripe connecting two disks in (a) is twisted, we have the

sphere in (b) exchanging the relations of heights between γ fi and γ& in

Fig. 4, (b).

4. Simply knotted spheres

Let S be a sphere in R * such that Γ(S*) consists of several disjoint

simple closed curves γf, • • • ,7*. We call such a projection S* a simple

projection. An equivalent class of spheres which contains the sphere S

is called a simply knotted sphere. It is obvious that a symmetric ribbon

sphere is a simply knotted sphere. We shall prove in this section that

every simply knotted sphere is equivalent to a symmetric ribbon sphere.

(4.1) 7* G > = 1» •• • > w ) is a trivial knot in R*.

Proof. Let @ be the unit sphere x?+y*+z*=l, and let h be a

homeomorphism of @ onto S. Put f=τth. Then f~\*/*) consists of two

disjoint circles cϊ an d cξ on @. The circles c] an d c* bound disks Dl



S IMP LY KNO TTE D S P H E R E S 139

an d Dl respectively on @ such that cl[\Dl =cl[\Dl =Q . Therefore, 7* is

the boundary of the singular disk Dl*=f(Dl), where singularities of

do not exist on 7?. Hence we can suppose that 7? is the boundary of

some non-singular disk by Deh's Lemma3). Therefore 7* is trivial. We

call Dί (£ = 1, 2) the interior of c£ .

(4. 2) 7? and 7* £/0 #0ί // w& homotopically for ι>

P r oof . Let f-\<γΐ)=cl[}cζ and /-X7?)=<£Lki. Let cϊ be an inner-

most one in these circles. Suppose that 7? links homotopically with γjf.

Then we have /(Z)ί)Γ|7ίΦ 0. Therefore γ j f is not a simple closed curve.

This contradicts the assumption.Let Jv be a parallel curve to cί sufficiently near to c{ on @. W ecan prove easily as above that /(r f

v) and 7?=/(£ί) do not link homo-

topically. Therefore, we have the following statement.

(4. 3) A ring-neighbourhood of ck

v, which does not contain another

inverse image of Γ(S*\ is mapped by f onto a topological non-twisted ring

surface in R\

N ow 7?'s are classified as fo l lows :(1) Both cl (k = l

92) are innermost in the circles of the inverse image

of Γ(S*).(2) O ne of c

k

v( f e = l , 2) is innermost and the other is not.

(3) Both cί are not innermost.

I f 7? is a sort of (2), we call it a canonical cutting.

(4.4) if both cί ( k =Ί9

2) are innermost, then 7? can be cancelled by

a deformation of S in R\

P r o o f . Since Dl and Dl do not contain the other circles of f ~ \Γ (S*) ) ,

they are mapped onto non-singular disks Dl*=f(Dl) and D ? * = / ( Z ) ? )respectively such that Di*n^;*=7? - Let 7 l =h(d) and 7?-%?). Sup-

pose that «(7?)<X7Ϊ) Ine first step, we deform a neighbourhood of

A(Z)?), so that the height of every point of h(Dl) is less than those ofpoints of f(Dl). During this deformation, S* is fixed. In the second step,

we deform h(Dl) parallel to R3, so that 7? may be cancelled. During the

deformation of the second step h(Dl) does not meet h(Dl).

(4.5) Let S be a simply knotted sphere. We can deform S into S ',

so that Γ(S'*) consists of canonical cuttings.

Proof . In virtue of (4. 4) we can suppose that Γ(S*) consists of

cuttings of (2) and (3). We shall prove by induction that non-canonicalcuttings (3) can be replaced by a finite number of canonical cuttings (2).

3) [4], p. 1 or [5], p. 223, theorem 1.



140 T. Y A J I M A

Let 7*, — 7?m

be non-canonical cuttings of S* in the kind of (3),an d let cl

ί9— c\

m, cl

mbe the circles of /-'(γ*) (i = l, — ,m ). Suppose

that c^ is an innermost one of these circles. We shall construct, in the

first step, a deformation of S* in R3, so that 7* may be replaced by

canonical cuttings.

Let E19

E2

be ring-neighbourhoods of c\y

c\ respectively, such that

each Ek

(k=l, 2) contains only the circle c j . In virtue of (4.3) we can

suppose that Ef=f(E1} is a ring surface on (x , jy)-plane and E$=f(E

2)

is a cylinder which is perpendicular to E% . Suppose that Dϊ* continues

to the inside of the ring surface E f .

The synopsis of the required deformation is illustrated in Fig. 6.

Fig. 6.

The shadowed part indicates that there exist some other parts of S*. If

was can prove that the intersection of the deformed cylinder ££* with

the shadowed part consists of canonical cuttings, then the proof is com-

plete. The detailed proof in the question is as follows (Fig. 7).

Let C be a component of /^-/"VXS*)) such that c\ may be a

member of boundaries of C. Let c , • • • , ςl7 c\> •••, c?A

be the rest ofboundaries of C, where 4/

sareinnermost circles of /-1(Γ(S*)) and c?/s

are not. For each c* . we take a parallel curve d\. to c?y

on C. If we

replace all c?y's by rf? y's, then w e have a domain Cj contained in C. Since

7*. is canonical, there exists a non-singular disk A*., which is parallel to

Dl

τ*=f(Dl

τ.) and BdA*.=f(d2

T.}. Thus we have a disk B* =C?U2£i*

UΣ^* such that jB*n(S-5Vl

)* consists of innermost circles on 5*.Let 5I f and £ * be two parallel surfaces to 5* illustrated as




Fig. 7.

broken lines in Fig. 7. The disks B lf and 5J* divide the cylinder E$

into three subcylinders E\*y

E\* and E\*y

where Blf and £?* are bounding

disks of E\* and E* respectively. Perforate the subcylinders E\* and

into j?2* an d JS| respectively, and connect these holes by a tube Γ*.Then we have the required cylinder E'

2* = (El*[)Blf)()(El*\jBϊ*)[]T*.

Thus the question is solved.

I n the last step of the proof, we shall show that the deformation in

R3 defined above is admissible as the projection of a deformation of S

in R\ It will be done very simply. Suppse that u(El) >u(E

2). D eform

h(E2\ so that th e height of every point of E I is less than that of the

points of S-h(El\ Then we can deform h(EΌ

2) parallel to R

3freely

f rom the other part of S. Thus, we have completed the proof.

Let S be a simply knotted sphere such that Γ(S*) consists of ca-nonical cuttings γf , • • • , γ*. Suppose that c ϊ , • • • , c\ c f, • • • , cl are circles

of /'VXS*)) on ©, where f-\tf)=cl\)<% (v = ly

— , n) and £Ϊ's are innermost

on ©. Hereafter we call el's circles of the first k ind, and el's are

those of the second kind. Let C 0, • • • , Cn

be the components of @— Σc?

Then each Cσ

(σ=0, 1, • • • , n) is mapped homeomorphically in 7? 3, but itmay happen that C

σis mapped not homeomorphically for some σ. A

cutting γί, such that cί and c? are contained in some Cσ, is called a s#//-

cutting of C*.

Let ^ j> ••• > be the bonding circles of Cσ. If C* contains no self-cutting, then it is obvious that M

σ=C

σ\]D\\] • • • []Dl

kis mapped into a

sphere in ί?3, where Z ) ϊ / s are the same as (4. 1). Therefore, the sphere



142 T. Y A J I M A

M* separates R3

into two parts. However, even in the case when C*

contains some self-cuttings, it is valid that M* separates R * into the

exterior and the interior of it.

(4. 6) Every simply knotted sphere S can be deformed into a sphere

S ', so that if the interior of M* of S'* contains some Cf's, then these C*'s

are mutually disjoint tubes and the relation of heights at two terminal

cuttings of a tube are mutually opposite for each T .

P ro o f . I n virtue of (4.5), we can suppose that Γ(S*) consists ofcanonical cuttings. Let C

σ, M

σbe as above. Let N* be a component of

S*—M* contained in the interior of M*. If all the vectors indicatingthe relations of heights at the cuttings of N* f| M* are either on N* or on

M*, then it is obvious that N* can be pulled out to the exterior of M* and

the cuttings of N*f}M* are cancelled. If N*f]M* consists of more than

two cuttings, and vectors of the relation of heights exist both on N*

an d M*, then take a sphere S% inside of M * such that th e interior of S%

contains only the cuttings on 2V*, an d pull out the interior of S% to theexterior of M*. The deformation in R

4corresponding to the pulling

out of 2V * m ay take place over or under A (C σ). I n consequence of this

deformation th e remaining part of 2V * in the interior o f M * splits into

the same number of tubes as the number of cuttings of N*(]M*.

Moreover, some of these tubes, each of which has the same relations of

heights at the terminal cuttings, can be pulled out cancelling these cuttings.

By repeating the above processes, we can get a spheres S ' which satisfies

the required conditions. Thus we have completed the proof.

Each C* has the orientation induced f rom that of S. According as

the vector indicating the orientation of C* points to the inside or to the

outside of M*, we say that C* is in the positive or in the negative situation

respectively.

N ow let us observe self-cuttings. Let c\ be an innermost one betweenthe circles of the second kind, and let C

0, Q be the components of

@ — Σ £ ? > such that C o Γ j C j — c ϊ , BdCQ=c\. Suppose that c\ is contained

in C0. Obviously C 0 — D\ is mapped onto a torus in R

3. According to

the situation of γf on the torus, there exist two cases (a) and (β\ where

7? in (Λ ) is homotopic to a longitude of the torus an d that in (β)

is homotopic to a meridian. Obviously C $ and C? in (a) are in mutually

opposite situations an d those of (β) are the same.

For a general case, let cl be the common boundary of Cλ

and Cμ,

and let γf be a sel f-cutt ing of C?=/(Cλ). In this case, there also existtwo cases as above. If C j f and C* are in the same situations, then we

call γf an admissible self-cuttig. If they are in opposite situations, then



S IMP LY K N O T T E D S P H E R E S 143

we call 7? a non-admissible self-cutting. It is easily proved that if 7?is an admissible self -cutting of C?, then C* is contained in the interior

of MJ, and that if γj is non-admissible, then C* is contained in the

exterior of MJ. Therefore, combining this with (4. 6), we have the

fo l lowing statement :

circle c\2

(4 . 7) // 7* is an admissible self-cutting of C j f , then there exists a

on Cλ

such that y?2 and γ v* are connected by a tube in the interior

of M$.

N ow we are going to prove that non-admissible self-cuttings can be

cancelled by a deformation of S in R\ We shall begin with a specialcase.

(4. 8) Let 7? be a non-admissible self -cutting of C J J S where c\ is an

innermost one between the circles of the second kind, and BdC0=d. Then

7? can be cancelled by a deformation of S in R4.

P ro o f . Let C , be the component of @-Σcξ such that C0Π C ι = cϊ.

Suppose that u(h(c\y)^>u(h(c^j) in Fig. 8 (tf j . Let Sl

be a sphere in R4

such that Sf contains C f f as the inside of it, and u(S^)^>u(h(C<$). Take

a small circle a on A (C 0) and a circle b o n S j respectively, and connect

these circles by a tube T in R4. Let ,4* and B* be disks on S* and onSf respectively such that they are contained in T*, and Bd A=a, BdB=b.

Take a concentric circle α' of a on the disk A. D e fo r m the inside A of

a', so that z^O^^OS— A ). Since St is trivial in R4—S, we can deform A

into the disk Sx

U T-B, so that ( - ) T and - (S β).

(03)

By this deformation, a new cutting parallel to each cutting of C?arises. But, in virtue of the definition of the height of S

ιythe pair of

cuttings 7?, 7ί* are cancelled as shown in Fig. 8 (<*3), where γί* is thenew cutting corresponding to 7?. Moreover, a pair of cuttings coorres-

ponding to one of the terminal cuttings of every tube contained in C?



144 T. Y A J I M A

is cancelled by the same reason as γf. Therefore, the cuttings on C*

are not changed as a whole except γf , and C 0 [ J C ι becomes a singlecomponent C Q . W e call th e deformation above th e reversing of C f.

Each circle cl of the second kind separates the sphere @ into two

disks. We shall distinguish one of these disks by Δvin the following

manner: Fix an arbitrary innermost circle cll

and put ΔVl

=Cσo

, whereBdC^= d . Let C

σιbe the neighbouring component of C

σQ. Let cl

2be

on e of the bounding circles of Cσι, which is distinct f rom c . Let Δv,be the disk, such that B d Δ ^,

2= cl

2and Δ V 2 ^ > Δ V l . For each circle cl.

contained in Δ V z , let Δv. be the disk such that Bd .=c

v. and Δ

v.dΔ

V2.

I n such a manner we can define a disk Δv for every cl, such that thecollection of disks {Δv} forms a directed set by the relation of inclusion.

(4.9) Let cl c\, • • • , c\m\ cl

l9• • • , cl

nbe the bounding circles of C

p.

Suppose that ΔλH D ( Σ Δ μ ^ L K Σ Δ V y ) and that 7?, γ^, • • • , γ*w are non-admis-

ι = l j = l

sible self-cuttings of C f . I f Δ^, • • • , Δ*m;Δ*ι ?• • - , Δ * M does not contain

any non-admissible self-cutting, then the sphere S can be deformed in R\

so that these non-admissible self-cuttings are cancelled.

Proof . P u t Δ(λ, y) = C p UΣΔvy. W e shall reverse the perforated

j = l

singular sphere Δ(λ, y)*. For this aim, we construct first of all a singular

sphere Sf which serves the same role as S? in (4. 8).

Let cl, CM (i = 1, •••, m ) be parallel curves to cj and c^ respectively

on Cp. Let C P

be the domain, such that BdCp

=cl[\^cl\J^cl.. Takeί=ι ' ;=ι

J

parallel disks Dl

μf to D J* and a parallel disk J5J* to D J * , so that 5 5 *

=/(?£) and BdDl*=f(ί$ respectively. Then Si*= ( C p * U / > i * U g / > i f ) U

ΣSJ. f o rm s a singular sphere, which does not contain any non-admissiblej

=ι J

self-cutting. Hence C*'s in Δ*. (j=ίy ~,n) and C * are in the samesituation. If these situations are negative, then deform Si* toward the

direction of the vector of the orientation of S*. If the situations of C*'sand Cf are positive, then deform S(* toward the opposite direction of

the vector. Thus we get a singular sphere Sf, which is similar to Si*.There occur three kinds of new cuttings as f o l l o w s :(1*) Cuttings which appear as the intersections of Sf and S*, where

tubes of S* run through Sf.(2*) Cuttings which appear as the intersections of Sf and S*, where

tubes of S f run through S*.

(3*) Cuttings which are intersections of Sf itself.Let γ* be an arbitrary cutting on ΔJ. Let γ*,, γ*β

and 7*β

be the

corresponding new cutttings to γ* of (1*), (2*) and (3*) respectively.



SIMPLY KNOTTED SPHERES 145

(1*)

(2*)

ς*

>1T

1

1<d > J

(3

Fig. 9.

We shall define the relations of heights at these cuttings as f o l l o w s :(1) Let 7'* be the other terminal cutting of the tube of S* in the

inter ior of S f. Then we define the height of S19

so that Sl

is both the

over surface at γf a and at γ(*.(2) The same relation as 7*, that is, if the tube of S, whose pro-

jection is parallel to the tube of Sf, is the over or under surface at 7*,

then the tube of S1

runs over or under S respectively.

(3) The same relation as 7 in the meaning of (2).

It is easily proved that Sl

is trivial in R4—S in virtue of (1), (2) and (3).

We shall reverse Δ*(λ, y), as we did in (4. 8). Take a small circle

a on h(Cp) and a circle b on Slβ Let A be the disk on h(C

p\ such that

BdA=a, and let B be the disk on Sl

such that BdB=b. Connect the

circles a, b by a tube T. Since S 1 is trivial in R4—S, we can deform thedisk A, so that A-*(S

1-B)(JT.

Thus we can reverse the perforated singular sphere Δ*(λ, v). But



146 T. Y A J I M A

it happens that γ?β

in (2) and γv*^ in (3) are not canonical cuttings. If

we replace these non-canonical cuttings with several canonical cuttings bythe method of (4. 5), then some new non-admissible self-cuttings may arise.

Therefore, we shall cancell these non-canonical self-cuttings by another

way.

For each Cσ

contained in Δλ—ΣΔμ,, take a point P

σon C

σ, so that

;=1

it is not contained in circles of the first kind. Connect Pσι

and Pσ2 by

a segment lσ 2

if and only if Cσι and Cσz

are mutually neighbouring

components, so that lσ 2

intersects the common boundary at a single

point, and that / σισ .2 does not meet circles of the fiirst kind. The unionof these segments makes a tree. We shall call the image of this tree

on S* a leading curve of the deformation.

The synopsis of the required deformation is explained as an enlar-gement of the tube T. We shall consider the most simple case, w = l.

Let Cσι

be the neighbouring component of Cp in Δ

Vl. Let (r, #, z) be a

cylindrical coordinate system of R3. We can suppose that A* and Cf — A*

are represented as the disk (r<l, z=ΐ) and the cylinder (r=l, 0<£<1)respectively. Therefore, C *

xis represented as the negative part of the

2-0

Fig. 10.




cylinder r=ί. Moreover, we can suppose that the corresponding part of

C*-A* in Sf is represented as the cylinder (r=2, 0<£<1) and thatthe tube T* is represented as the ring surface (l r 2, 2 = 1). Suppose

that /*σ

is represented as the segment (r = 1,0 = 0, —1<^2^1), where

P? = (l,0, 1) and P*t= (l, 0, -1).

Take points Oί = (l, -00> 1) and O f =(1, 0

0,1) on α* and J ? f = ( 2 , -0

0>1)

and |?f =(2, # 0, 1) on ft*, where 00

is a sufficiently small positive number.

Let Oί*, #ί* Q£*, /?5* be the corresponding points of O f, Rf O f, £frespectively on the plane z=—l.

I n virtue of (2), it is easily proved that the surface composed of the

disks Q 1Q 2Q/

2Q ίy Q1Q2R2R1

a

nd R{R'JRJt^ is deformed into the surface com-posed of the disks Qfί(R(R

ιyQ{Q'Jt 'Jt{ and Q

2QίRίR

2. In consequence

of this deformation, the pair of cuttings 7*χ

and γ β becomes a single

canonical cutting.

I n the case of n^>l9

we can also enlarge the tube T in the similar

way as above, so that non-canonical cuttings γ*β, • • • ,7ίwβ

are cancelled.

Therefore, even in the case when the leading curve has branched points,

we can continue the deformation as far as all non-canonical cuttings are

not cancelled.

(4.10) Suppose that y j f in (4. 9) is an admissible self-cutting. In this

case, we can also cancel non-admissible self-cuttings in Δ*.P r oof . First of all we reverse the parforated singular sphere Δ*(λ, y)

in th e same method as (4. 9 ). In consequence of the deformation, there

happen a new non-admissible self-cutt ing as the intersection of C f andSf. But this is the case of m=0 in (4, 9). Therefore we can cancel it

by the reversing of the deformed sphere.

In virtue of (4. 8), (4. 9) and (4.10), we can prove by induction that

non-admissible self-cuttings of S* can be cancelled. Thus we have the

f o l l o w i n g :(4.11) Every simply knotted sphere can be deformed, so that the

deformed sphere does not contain any non-admissible self-cutting.

N ow we shall prove the fo l lowing theorm, which is the purpose of

this section.

(4.12) Theorem 1. Every simply knotted sphere is equivalent to a

symmetric ribbon sphere.

P r o o f . From (4.6), (4.7) and (4.11), we can suppose that the circles

of the first kind on every Cσ

are divided into several pairs, so that the

circles in a pair are mapped into th e pair of terminal cuttings of a tubein M*. Moreover, we can suppose that these terminal cuttings have an

opposite relations in height.



148 T. Y A J I M A

Renominate the circles of /"'(/XS*)), so that (cl>iy

c^}\ (i = l, • • • ,A»)is a couple on Cσ, where cifί

is mapped on the over surface at γ * f < and

c£fϊ is mapped on the under surface at γ^f.

Let Pσo

(<r=0, 1, • • • , n) be a point on Cσ

which is not contained in

the circles of the first kind, and let Pσi

be a point on c* , (i = 1, • • • ,/(σ*)).

Connect Pσ < with Pσ><+1

by a simple arc s?f<+1

(ι = 0, ••• ,/(σ)-l), so that

these arcs are mutually disjoint and do not intersect f~l(Γ(S*)) except

their boundaries. Put Lσ=( Σ 5?ί+1)U(Σ °σ ί), where D J , is the interior

f= 0 ' /=!

of the circle < £> f

. Moreover connect Pσo

and Pτ o by a simple arc /στ

if

and only if Cσ and Cτ are mutually neighbouring components, so that /στ

intersects the common boundary at a single point and does not intersect

Lσand Lτ

except its end points.

F i g . 11.

Let N be a sufficiently small neighbourhood of ΣLσ

U Σ A π Then

th e bounding curve k of N satisfies th e conditions

(1) K separates @ into two disks D+ and D _ , where D+ contains

clti

an d D _ contains cf

σ\

{( o - = 0, • • -, « ; / = !,••• ,/(σ)),

(2) & intersects each < £ , • at two points.

Put £*=/(£) and deform S, so that u(h(k}) =0, without changing the rela-

tion of heights at every cutting. Then D+ and Z ) _ satisfies the conditionsof the inverse image of a ribbon. It is obvious that h(D+) and λ(Zλ.) are

symmetric with respect to R\ Thus we have completed the proof.




5. The Alexander polynomials

We are now in a position to prove the existence of a sphere which

is distinct f rom simply knotted spheres. But before the discussion we

shall review some history concerning knotted spheres.

A f t e r papers concerning spinning spheres by E. Artin [1] and others

[2], [3], R. H. Fox and J. W. Milnor [6] discussed about slice knots or

null-equivalent knots, which appear as an intersection of knotted spheres

with a 3-dimensional subspace. In their paper they proved that the

Alexander polynomials of slice knots are given in the form o f f ( t ) f ( t ~l) .

Several years later H. Terasaka [8] proved the converse of Fox-Milnor'stheorem. In his paper, he tried to give an alternating proof of Fox-

Milnor's theorem. But, since he assumed that every null-equivalent knot

is represented as a ribbon knot, there exist some gap4)

in the proof.

However it is valid that the Alexander polynomial of a ribbon knot is

given in a fo r m f ( t ) f ( t ~1} as shown in Terasaka's paper. Following

Terasaka's work, S. Kinoshita [9] proved the existence of a knotted

sphere which has the given Alexander polynomial /(/) with /(1)=±1.For our purpose nothing new is necessary, but only a remark to Terasaka's

alternating proof of Fox-Milnor's theorem is sufficient.Let A be an arbitrary knot represented by a regular projection on

the ground plane, and let C be a trivial knot represented by a circle

which is disjoint f rom A on the ground plane. Connect a small arc a

of A to a small arc γ of C by a band B which is represented by a pair

of parallel curves, so that (A\]B\]C)-(<*{]*/) forms a knot & , where B

may tangle with A, C or B itself. It is obvious that if A is a ribbon

knot, then k is also a ribbon knot, and that conversely every ribbon knot

is given in such a fashion.

We call the boundary curve of B, whose orientation coincides withthe direction of B, that is f r o m A to C, the positive side of B, and call

the opposite one the negative side. By an under crossing of B with A

or C, B splits into several parts Bιy

• • • ,Bn

in this order. Let bioy

• .• , ίf. /c/)

be the parts of the positive side of Bf

(ί = l, • • • , n), which do not contain

under crossing arcs at intersections of B itself, and let bik

(k=09

•••, /(/))

be the corresponding negative side of b{ k. A and C are also divided by

B into A > •'•ΆN

and Q, — , CM

respectively. Moreover A{

(ι = l, •••, N)

may splits into aiΛ, • • • , 0

i>nCi)by some crossing points of A itself (Fig. 12).

Using the same notation for generating elements of §(&) as arcs of

k , Terasaka did as follows:

4) [7], p. 173, Problem 25.



150 T. Y A J I M A

Fig. 12.

First of all he remarked that we can set

(5. 1) δf.Λ o = — = δ ίf «•• , n)

in the computation of the Alexander matrix of k. However, it should

be noticed that the relations (5. 1) are not valid in the group f$(£). In

the next step, replacing generators 6 ί > 0 , — , &;,/a)

— bn>0

, • • • , bn /Cl0

byB

19— , β

Man d &

M, — ,£

M(υ— £ „ , < > > — > *»,«»> and eliminating £

M, • • • , /(1)

b2ι,~ ,b

2/ C 2 ) — *» ι > • • • > * » / c * ) - ι > he got the fo l lowing Alexander matrix

MM:

llβ(*)ll

*

*

*

1 °1 I 1 1 / 0 0 1 1

0

i o

0

*II WII

0

0

*

0

IM000II

I n the above matrix, the rows which contain the square matrix

||£(#)|| of order n correspond to the relations at under crossings of the

band B with A or C, and the rows which contain the square matrix

11/0011 of order n correspond to the relations along the positive side of

B. The minor | |ΔXjc) | | is the Alexander matrix for the knot A, and| | Δ c O O | | is the Alexander matrix for the trivial knot C. In the last stephe proved that g(x)= ±x

mf(x~

1\ Thus we have




(5. 2) // k is a ribbon knot, then there exists a polynomial f(x\ such

that Δk(x}=±xmf(x)f(x~l\ where Δ

k(% ) is the Alexander polynomial of k.

N ow we shall prove the fo l lowing :

(5. 3) Theorem 2. Let k and S be the ribbon knot and the symmetric

ribbon sphere of an arbitrary ribbon τ. Then, there exists a polynomial

f(t\ such that

P r oof . First of all, we shall prove the most simple case, that is the

case where the knot A in Terasaka's proof is trivial. In virtue of the

construction of S* in § 3, S(S) has the generators :

<Kft)=

<Kft,o)= ••• =

=σ<A)=-==0"(ft,o)=••'=

= ••• = cr(CM) ,

each of which corresponds to the part of k represented by the notationin parenthesis.

N ow we shall define a mapping φ of the generator system of

onto the generator system of S(S), such that

φ(bitk

} = φ(bi> k

) =

Then we have easily that

(i = 1, ••• , n)

=-=α-(CM),

an d that th e mapping φ forms a homomorphism of $(k) ontoObviously the group §(S) is isomorphic to the group τ$*(k\ which has the

same generators as §(&) and has the same relations as that of % ( k ) plus

the fol lowing relations :

(*) A = ft =••• =£. =!.

We add the new relations (*) in Mk

to compute the Alexander matrix

of 8*(&). Then we have Δs(0=/(0 The general case of the theorem

is proved easily by induction.(5. 4) Corollary. Every symmetric ribbon sphere has the Alexander

polynomial.



152 T. Y A J I M A

Since the elementary idea!5) of the Alexander matrix of the Fox's

sphere is not principal, that is, it has not the Alexander polynomial,

combining (5. 4) with Theorem 1, 2, we have the following main theorem

of this paper:

(5. 5) Theorem 3. There exist spheres which are distinct from simply

knotted spheres.

OSAKA CITY U N I V E R S I T Y

Bibliography

[1] E. Artin : Zu r Isotopie zweidimensionaler Flάchen in R4, Abh. Math. Sem.

Univ. Hamburg 4 (1925), 174-177.[2] E. R. Van Kampen: Zur Isotopie zweidimensionaler Flάchen in R± , Abh.

Math. Sem. Univ. Hamburg 6 (1927), 216.

[3] J. J. Andrews and N. L. Curtis: Knotted2-spheres in the 4-sphere, Ann.of Math. 70 (1956), 565-571.

[ 4 ] C. D. Papakyriakopoulos: On Dehrfs lemma and the asphericity of knots,Ann. of Math. 66 (1957), 1-26.

[ 5 ] T. Homma : On Dehn's Lemma for S3, Yokohama Math. J. 5 (1957), 223-

244.[ 6 ] R. H. Fox and J. W. Milnor: Singularities of 2-spheres in 4-space and

equivalence of knots, unpublished. C f. Bull. Amer. Math. Soc. 63 (1957),

406.[ 7 ] R. H. Fox: A quick trip through knot theory, Top. of 3-manifolds and

related topics, edited by M. K. Fort Jr., Prentice Hall (1962).

[ 8 ] H. Terasaka: On Null-Equivalent Knots, Osaka Math. J. 11, No. 1, (1959).

[ 9 ] S. Kinoshita : On the Alexander polynomials of 2-sphere, Ann. of Math.74 (1961), 518-531.

[10] T. Y a j i m a : On the fundamental groups of knotted 2-manifolds in the Δ-space,

J. of Math., Osaka City Univ. 13 (1962), 63-71.

5) [7], p. 127.

takeshi yajima- on simply knotted spheres in r^4

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